Question

question
using a calculator or statistical software, find the linear regression line for the data in the table below.
enter your answer in the form $y = mx + b$, with $m$ and $b$ both rounded to two decimal places.

x	y
1	2.19
2	1.92
3	2.79
4	3.81
5	4.72

Explanation:

Step1: Identify the data points

We have the following data points: \((0, 2.12)\), \((1, 2.19)\), \((2, 1.92)\), \((3, 2.79)\), \((4, 3.81)\), \((5, 4.72)\)

Step2: Use linear regression formula (or calculator/software)

The general form of a linear regression line is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. Using a calculator or statistical software (e.g., using the formula for slope \(m=\frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}\) and \(b=\frac{\sum y - m\sum x}{n}\) where \(n = 6\)):

First, calculate the necessary sums:

• \(\sum x = 0 + 1 + 2 + 3 + 4 + 5 = 15\)
• \(\sum y = 2.12 + 2.19 + 1.92 + 2.79 + 3.81 + 4.72 = 17.55\)
• \(\sum xy = (0\times2.12)+(1\times2.19)+(2\times1.92)+(3\times2.79)+(4\times3.81)+(5\times4.72)=0 + 2.19 + 3.84 + 8.37 + 15.24 + 23.6 = 53.24\)
• \(\sum x^2 = 0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 0 + 1 + 4 + 9 + 16 + 25 = 55\)

Now, calculate \(m\):

$$ m=\frac{6\times53.24 - 15\times17.55}{6\times55 - 15^2}=\frac{319.44 - 263.25}{330 - 225}=\frac{56.19}{105}\approx0.5351 $$

Rounding \(m\) to two decimal places, \(m\approx0.54\)

Next, calculate \(b\):

$$ b=\frac{17.55 - 0.5351\times15}{6}=\frac{17.55 - 8.0265}{6}=\frac{9.5235}{6}\approx1.5873 $$

Rounding \(b\) to two decimal places, \(b\approx1.59\)

Answer:

\(y = 0.54x + 1.59\)